Existence of Radius of Convergence of Complex Power Series/Absolute Convergence

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Theorem

Let $\xi \in \C$.

Let $\displaystyle S \paren z = \sum_{n \mathop = 0}^\infty a_n \paren {z - \xi}^n $ be a complex power series about $\xi$.

Let $R$ be the radius of convergence of $S \paren z$.


Let $B_R \paren \xi$ denote the open $R$-ball of $\xi$.

Let $z \in B_R \paren \xi$.

Then $S \paren z$ converges absolutely.


If $R = +\infty$, we define $B_R \paren \xi = \C$.


Proof

Let $z \in B_R \paren \xi$.

By definition of the open $R$-ball of $\xi$:

$\cmod {z - \xi} < R$

where $\cmod z$ denotes the complex modulus of $z$.

By definition of radius of convergence, it follows that $S \paren z$ converges.


Suppose $R$ is finite.

Let $\epsilon = R - \cmod {z - \xi} > 0$.

Now, let $w \in B_R \paren \xi$ be a complex number such that $R - \cmod {w - \xi} = \dfrac \epsilon 2$.

Such a $w$ exists because, if at a loss, we can always let $w = \xi + R - \dfrac \epsilon 2$.

If $R = +\infty$, then let $w \in C$ be any complex number such that $\cmod {z - \xi} < \cmod {w - \xi}$.

Then:

\(\displaystyle \limsup_{n \mathop \to \infty} \cmod {a_n \paren {z - \xi}^n}^{1/n}\) \(<\) \(\displaystyle \limsup_{n \mathop \to \infty} \cmod {a_n \paren {w - \xi}^n}^{1/n}\) as $\cmod {z - \xi} < \cmod {w - \xi}$, we have $\cmod {z - \xi}^n < \cmod {w - \xi}^n$
\(\displaystyle \) \(\le\) \(\displaystyle 1\) $n$th Root Test: $S \paren w$ is not divergent

From the $n$th Root Test, it follows that $S \paren z$ converges absolutely.

$\blacksquare$


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